Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{-4k + 36}{k^2 + 5k + 4} \div \dfrac{9k - 81}{-9k - 9} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4k + 36}{k^2 + 5k + 4} \times \dfrac{-9k - 9}{9k - 81} $ First factor the quadratic. $q = \dfrac{-4k + 36}{(k + 1)(k + 4)} \times \dfrac{-9k - 9}{9k - 81} $ Then factor out any other terms. $q = \dfrac{-4(k - 9)}{(k + 1)(k + 4)} \times \dfrac{-9(k + 1)}{9(k - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -4(k - 9) \times -9(k + 1) } { (k + 1)(k + 4) \times 9(k - 9) } $ $q = \dfrac{ 36(k - 9)(k + 1)}{ 9(k + 1)(k + 4)(k - 9)} $ Notice that $(k - 9)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 36(k - 9)\cancel{(k + 1)}}{ 9\cancel{(k + 1)}(k + 4)(k - 9)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac{ 36\cancel{(k - 9)}\cancel{(k + 1)}}{ 9\cancel{(k + 1)}(k + 4)\cancel{(k - 9)}} $ We are dividing by $k - 9$ , so $k - 9 \neq 0$ Therefore, $k \neq 9$ $q = \dfrac{36}{9(k + 4)} $ $q = \dfrac{4}{k + 4} ; \space k \neq -1 ; \space k \neq 9 $